1,141 reputation
212
bio website
location
age
visits member for 2 years, 10 months
seen Sep 22 at 22:03

Jan
21
comment List of connected Lie subgroups of $\mathrm{SL}(2,\mathbb{C})$.
Sorry I should have precised I am only interested in Lie subgroups of real dimension at least one.
Jan
21
asked List of connected Lie subgroups of $\mathrm{SL}(2,\mathbb{C})$.
Dec
30
asked Difference between diffeomorphisms fixing a point or a whole neighborhood.
Dec
29
accepted Exact sequences and (semi) direct product
Dec
29
comment Exact sequences and (semi) direct product
Thanks. A hint for point 3. ?
Dec
29
asked Exact sequences and (semi) direct product
Dec
25
comment Geodesic flow on a manifold with negative curvature is ergodic
If you are interested in the proof of Mostow rigidity, you can certainly avoid using this result.
Dec
19
revised Linear group action over an hermitian space.
added 453 characters in body
Dec
19
asked Linear group action over an hermitian space.
Dec
14
comment Closed but not exact one-form on $S^2$
Ok I read a bit fast your hypothesis and missed the fact that you removed 3 points. Whatever, I'm not sure I understand what you're asking for in your comment. Could you precise a bit ?
Dec
14
comment Closed but not exact one-form on $S^2$
This cannot happen on $S^2$ because the non-exactness of a closed form must come from a hole on the surface (in rough words). You should try to prove the following : let $S$ be a surface which is simply connected(this express the fact that $S$ has no hole), then every closed $1$-form is exact.
Dec
14
comment Is the matrix defined by $\bar{K}_{ij}=f(x_i)f(x_j)$ for a real valued function $f$ semi-positive-definite?
Other remark, such a matrix will always be of rank at most $1$, since all lines are colinear to $(f(x_1),...f(x_n))$. It implies that it will never be definite positive provided that $n \geq 2$.
Dec
14
comment Is the matrix defined by $\bar{K}_{ij}=f(x_i)f(x_j)$ for a real valued function $f$ semi-positive-definite?
First of all, put $f \equiv 0$ to get a matrix which is not positive-definite since $\overline{K} = 0$ .
Dec
11
comment Let $A$ be an abelian group. Show that $\mathrm{Hom}(\mathbb Z, A)$ is isomorphic to $A$.
Assuming $Z$ is the set of integers, I invite you to check that $ \varphi \in \mathrm{Hom}(\mathbb{Z}, A) \longmapsto \varphi(1)$ is the isomorphism you are seeking.
Dec
11
comment When does an element of $\mathrm{Sl}_n(\mathbb{R})$ preserve a scalar product?
I'm not sure. Here you are using the fact that complex eigenvalues are conjugates, and for each pair of conjugate you use only one eigenvector and take the real and imaginary part. DO you see what I mean ?
Dec
11
comment When does an element of $\mathrm{Sl}_n(\mathbb{R})$ preserve a scalar product?
Ok I understand how you want to proceed but now we should check that $e_i,x_j,y_j$ actually form a basis for $\mathbb{R}^n$. I think that using computation you used to prove that $x_j$ and $y_j$ are linearly independent should work but I doesn't seem obvious.
Dec
11
comment When does an element of $\mathrm{Sl}_n(\mathbb{R})$ preserve a scalar product?
Actually I don't really understand how you find the $f_j$ and $g_j$. Precisely, why do their coefficients belong to $\mathbb{R}$ since the diagonalization ii over $\mathbb{C}$ ?
Dec
11
comment When does an element of $\mathrm{Sl}_n(\mathbb{R})$ preserve a scalar product?
Yes exactly. I'm not sure how to be clearer :)
Dec
11
revised When does an element of $\mathrm{Sl}_n(\mathbb{R})$ preserve a scalar product?
added 8 characters in body
Dec
11
comment When does an element of $\mathrm{Sl}_n(\mathbb{R})$ preserve a scalar product?
For example any conjugate of a rotation preserve a scalar product, but does not belong to $O_2(\mathbb{R})$.